If minimum value of term free from $x$for

${\left(\frac{x}{\sin \theta }+\frac{1}{x\cos \theta }\right)}^{16}$is $L_1$ in $\theta \in \left[\frac{\pi }{8},\frac{\pi }{4}\right]$ and $L_2$ in $\theta \in \left[\frac{\pi }{16},\frac{\pi }{8}\right],$then

$\frac{L_2}{L_1}=$

The general term in the expansion of

${\left(\frac{x}{\sin \theta }+\frac{1}{x\cos \theta }\right)}^{16}$is

$\begin{array}{l} T_{r+1}{=}^{16}{C}_r{\left(\frac{x}{\sin \theta }\right)}^{16-r}{\left(\frac{1}{x\cos \theta }\right)}^r\\ \Rightarrow T_{r+1}{=}^{16}{C}_r{\left(\frac{1}{\sin \theta }\right)}^{16-r}{\left(\frac{1}{\cos \theta }\right)}^r{x}^{16-2r}\end{array}$

For this to be independent of $x$, we must have 

$16 - 2r = 0 => r = 8.$

Putting $r = 8$, we obtain 

$\begin{array}{l} T_9{=}^{16}{C}_8{\left(\frac{1}{\sin \theta }\right)}^8{\left(\frac{1}{\cos \theta }\right)}^8\\ \Rightarrow T_9{=}^{16}{C}_8{\left(\frac{1}{\sin \theta \cos \theta }\right)}^8{=}^{16}{C}_8\times 2^8\frac{1}{{\sin }^{8}2\theta }\end{array}$

For $\theta \in \left[\frac{\pi }{8},\frac{\pi }{4}\right],{\sin }^{8}2\theta$ 0 is maximum at $\theta =\frac{\pi }{4}$ and hence $T_9$ is

minimum at $\theta =\frac{\pi }{4}$

$\therefore L_2{=}^{16}{C}_8\times 2^8\times \frac{1}{{\sin }^8\frac{\pi }{4}}{=}^{16}{C}_8\times 2^8\times (\sqrt{2}{)}^8{=}^{16}{C}_8\times 2^{12}$

Hence $\frac{L_2}{L_1}=\frac{{ }^{16}{C}_8\times 2^{12}}{{ }^{16}{C}_8\times 2^8}=2^4=16$